Question

# In figure, the coefficient of static friction between the floor and block B is 0.30. The coefficient of static friction between the block B and A is 0.15. The mass of block A is m2 kg and B is m kg. What is the maximum horizontal force F in Newtons that can be applied to block B, so that the two blocks move together?

A
1.35 mg
B
0.45 mg
C
0.675 mg
D
0.9 mg

Solution

## The correct option is C 0.675 mgmA=m/2 kg,  mB=m kg μA=0.15,    μB=0.3 ⇒Let both the blocks are moving together with common acceleration a. The maximum force will correspond to the maximum value of f1, which can sustain the required acceleration for block A From FBD of block, maximum value of f1=μANA=μA=μamAg   ∴a=μAmAgmA=μAg=0.15g m/s2 ...(i) Now, on the system of blocks A+B, applying equation of dynamics from the FBD: Maximum value of f2=μBN=μB(mA+mB)g Hence applying newton's 2nd law,    ⇒F−μB(mB+mA)g=(mB+mA)a putting a=0.15g from Eq. (i) F=0.3(mB+mA)g+(mB+mA)0.15g ⇒F=0.45(mB+mA)g  F=(m+m2)0.45g ∴F=32m×0.45g=0.675 mg NPhysics

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